@19rsb c is the correct ans
@gupanki2 said:@scrabbler Yep answer is correct...did u try hit and trial method or any specific logic/equation , if so, kindly share the same.
Very much hit and trial, same starting point (average in 50s) as @SarayuSheshadri pointed out
regards
scrabbler
regards
scrabbler
@gupanki2 said:A teacher asked a student to add nine two digit number such that each number is tem more than previous number. But the student has all the numbers, which are formed by interchanging the digits of each of the given number. Surprisingly, the sum he got was the same as the teacher had asked for. Find the differnce of the digits in the least of the numbers the student was asked for to add?
The no.s should be of the form 1x, 2x, 3x,...9x (i.e. 10+x, 20+x...etc) and the reverse should be x1,x2,x3,...x9.
Sum of 1st set = 10*(1+2+....+9) + (9*x) = 450 + 9*x -> (1)
Sum of reverse set = 10*9*x + (1+2+...+9) = 90*x + 45 -> (2)
(1)= (2) => 81*x = 405
=> x=5
So min. no. = 15 and diff. = 4.
Sum of 1st set = 10*(1+2+....+9) + (9*x) = 450 + 9*x -> (1)
Sum of reverse set = 10*9*x + (1+2+...+9) = 90*x + 45 -> (2)
(1)= (2) => 81*x = 405
=> x=5
So min. no. = 15 and diff. = 4.
@meow14 cause any MIDDLE number has to be in fifties ...as starting number has to be in 10-19 and ending number in 90-99...@SarayuSheshadri correct me if i am wrong
@gupanki2 said:@meow14 cause any MIDDLE number has to be in fifties ...as starting number has to be in 10-19 and ending number in 90-99...@SarayuSheshadri correct me if i am wrong
Yep.. Exactly the reason..
@SarayuSheshadri yaaar even i concluded it intially dat middle number has to be betwn 50-59 but iss step se answer nahi conclude kr paya....afta ya said the average wali baat then laga ki us step se in a go nikal jaega answr
@janta.... there are 2 villages B and C and there head count is compiled....B has 5114 more males than C.....C has twice as many females as males.... B has 3004 more males than females... B has 9118 females fewer than those of C.... Find no. of females @ B???
**Edited**
@gupanki2 said:@janta.... there are 2 villages B and C and there head count is compiled....B has 5114 more males than C.....C has twice as many females as males.... B has 3004 more males than females... B has 9118 females fewer than those of C.... Find no. of females @ B???
Bm - Cm = 5114..(1)
Cf = 2*Cm..(2)
Bm -Bf = 3004..(3)
Cf - Bf = 9118..(4)
Using (1) ,(2),(3)
Bf - Cm = 2110 => -Cf/2 + Bf = 2110 => -Cf + 2*Bf = 4220...(5)
Using (4), (5)
Cf = (2*9118 +4220) = 22456
Bf = Cf - 9118 = 13338 ?
PS: There could be calculation mistakes :|
Total Number of 4-digit numbers which contain not more than 2 different digits:
1) 783
2) 776
3) 576
4) 496
5) None of these
PS: No OA 😃
@soham2208 said:Total Number of 4-digit numbers which contain not more than 2 different digits:1) 7832) 7763) 5764) 4965) None of thesePS: No OA
There will be 3 cases for this where we have no distinct digit, 1 distinct and 2 distinct.
No distinct digit : Let the number be aaaa=9 ways
1 distinct : abbb = 4c3*9*9, a in 9 ways and b in 9 ways
2 distinct : aabb= position of a can be selected in 4c2 ways.
a can be written in 9 ways and b in 9 ways
Total = 4c2*9*9 but each number is taken twice so, 4c2*9*9/2
Adding all the above totals,
We get 576 ways
We get 576 ways
@soham2208 said:Total Number of 4-digit numbers which contain not more than 2 different digits:1) 7832) 7763) 5764) 4965) None of thesePS: No OA
all same diff digit + 2 diff digit
[9] + [9+9C2*2*4 + 9C2*4!/2!*2!] + 27 + 27 = 576
If n is a natural number, then what is the sum of all the possible distinct remainders when
8^n + 7^n + 3^n + 2^n is divided by 10?
8^n + 7^n + 3^n + 2^n is divided by 10?
(a) 6
(b) 10
(c) 2
(d) 0
@soumitrabengeri said:If n is a natural number, then what is the sum of all the possible distinct remainders when8^n + 7^n + 3^n + 2^n is divided by 10?(a) 6 (b) 10(c) 2 (d) 0
when n=odd it is divisible by 10
n=2
64+49+9+4 = x6
n=4
6+1+1+6 = x4
10 is answer clearly.
The function f(x, y), defined on the set of ordered pairs of positive integers, satisfies the following properties.
f(x, x) = x
y — f(x, y) = x — f(y, x)
(x + y)f(x, y) = yf(x, x + y).
Find f(14, 52).
OPTIONS
1) 378
2) 456
3) 364
4) 728
5) None of these
f(x, x) = x
y — f(x, y) = x — f(y, x)
(x + y)f(x, y) = yf(x, x + y).
Find f(14, 52).
OPTIONS
1) 378
2) 456
3) 364
4) 728
5) None of these
The average marks of section A students are 20% less than the average marks of section B students. On the other hand, the total marks of section A are 20% more than the total marks of section B. Thus, the number of students in section B is what percent of the number of students in section A?
OPTIONS
1) 4%
2) 150%
3) 25%
4) None of these
OPTIONS
1) 4%
2) 150%
3) 25%
4) None of these
@sujamait said:The average marks of section A students are 20% less than the average marks of section B students. On the other hand, the total marks of section A are 20% more than the total marks of section B. Thus, the number of students in section B is what percent of the number of students in section A?OPTIONS1) 4% 2) 150% 3) 25% 4) None of these
Taking an example,
Let the avg marks of section B be 15
Then avg marks of section A will be 12
Total marks of A > Total marks of B by 20%
This is possible when Total(A) = 108 and Total(B) = 90
Hence, no of students (A) = 9 and no of students(B) = 6
6/9 = 66.66%
Clearly, the answer is none of these